Short Answer Type
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y – cos y = x ...(1)
Differentiate both sides w.r.t. x, we get,
y' + sin y . y' = 1
⇒ y'(1 + sin y) = 1 ...(2)
L.H.S. = (y sin y + cos y + x) y'
= (y sin y + y) y' [∵ of (1)]
= y (1 + sin y) y' = y. 1 [∵ of(2)]
= y
= R.H.S.
∴ y – cos y = x is a solution of (y sin y + cos y + x) y' = y.
In the following, verify that the given functions, (explicit or implicit) is a solution of the corresponding differential equation:
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